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Sindrei [870]
3 years ago
8

Polarizing windows, filters, etc. are often used to reduce the amount of light that enters the lens of a camera or into a room o

r a car. A library atrium has an overhead skylight that lets in too much light during the day which heats up the interior of the library far too much. The building engineer installs new double paned polarizing sky lights to reduce the intensity. If sunlight, which is unpolarized, has an average intensity of 1250 W/m^2.
Required:
What angle should the polarizing axis of the second pane of the window make with the polarizing axis of the first pane of the window in order to reduce the intensity of the sunlight to 33% of the original value?
Physics
1 answer:
LekaFEV [45]3 years ago
7 0

Answer:

The answer is "35.6^{\circ}"

Explanation:

The sunlight level of the first panel:

I_1 = \frac{I_o}{2}

When the light of this intensity passes through the second window:

I_2 = I_1 \cos^2 \theta\\\\I_2 = \frac{I_o}{2} \cos^2 \theta

 \frac{I_2}{I_o} = 0.33 (33\%) \\\\

therefore,  

0.33 = \frac{1}{2} \cos^2 \theta\\\\\cos^2 \theta = 0.66\\\\\cos \theta = \sqrt{0.66} = 0.8124\\\\\theta = \cos^{-1}( 0.8124) = 35.6^{\circ}\\\\

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antiseptic1488 [7]

Answer:

A super hero because most people hate villains and I don't wanna be hated :)

4 0
3 years ago
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A 27 kg disk with a radius of 1.3m is spinning at an angular speed of 15 rad/s. What is the rotational kinetic energy of the dis
satela [25.4K]

Answer:

the rotational kinetic energy of the disk is 5,133.375 J

Explanation:

Given;

mass of the disk, m = 27 kg

radius of the disk, r = 1.3 m

angular speed, ω = 15 rad/s

The rotational kinetic energy of the disk is calculated as;

K.E_{rot} = \frac{1}{2}I \omega^2\\\\  where;\\I \ is \ moment \ of \ inertia\\\\K.E_{rot} =  \frac{1}{2} \times (mr^2) \times \omega ^2\\\\ K.E_{rot} = \frac{1}{2} \times (27\times 1.3^2) \times \ 15^2\\\\K.E_{rot} = 5,133.375 \ J

Therefore, the rotational kinetic energy of the disk is 5,133.375 J

4 0
3 years ago
What is the distance from the moon to the point between Earth and the Moon where the gravitational pulls of Earth and Moon are e
Ugo [173]

Answer:

the point between Earth and the Moon where the gravitational pulls of Earth and Moon are equal is <em>E)3.45 × 10⁸ m</em>

Explanation:

The force that the Earth exerts on a mass m is

F_e = (G M_e m) / R_e²

where

  • G is the universal gravitational constant
  • M_e is the mass of Earth
  • R_e is the radius of Earth

The force that the Moon exerts on a mass m is

F_m = (G M_m m) / R_m²

where

  • G is the universal gravitational constant
  • M_m is the mass of the Moon
  • R_m is the radius of the Moon

Therefore, the point where the gravitational pulls of Earth and Moon are equal is:

F_e = F_m

R_e + R_m = R = 3.84×10⁸ m

Thus,

(G M_e m) / R_e² = (G M_m m) / R_m²

M_e / R_e² = M_m / (R - R_e²)

(R - R_e²) / R_e² = M_m / M_e

(R - R_e) / R_e = (M_m / M_e)^1/2

R_e(R/R_e -1) / R_e = (M_m / M_e)^1/2

R/ R_e = (M_m / M_e)^1/2 + 1

R_e = R / [(M_m / M_e)^1/2 + 1]

R_e = (3.84×10⁸ m) / [(7.35 x 10²² kg / 5.97 x 10²⁴ kg )^1/2 + 1]

R_e = 3.45 × 10⁸ m

Therefore, the  point between Earth and the Moon where the gravitational pulls of Earth and Moon are equal is <em>3.45 × 10⁸ m.</em>

4 0
3 years ago
Calculate the velocity displace = 3500km and time = 3 hours
Alisiya [41]
Divide 3500km by 3h (according to the equation v = d/t) and you get 1167km/h. 
7 0
3 years ago
An object initially traveling at a velocity of 52 m/s experiences an acceleration of 9.8 m/s^2 how much time will it take to com
Goshia [24]

Answer:

5.3 seconds

Explanation:

Vx = Vx0 + Ax .T

A= de-acceleration

Vx =0   final at rest velocity

Vx0 = initial velocity

T= time to bring the object to a stop

therefore

0 = 52 m/s - (9.8 m/s^2) . T

-52 m/s/ -9.8 m/sec^2 =T

5.3 seconds= T

8 0
3 years ago
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